In Problems 76, 77, 78, and 79 you are given the equation(s) used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation(s).

1/2(0.30 kg)(0 m/s)² + 1/2(3.0 N/m)(Δx2)²

= 1/2(0.30 kg)(v1x)² + 1/2(3.0 N/m)(0 m)²

Video transcript

Hey, everyone in this problem, we're given the equation one half multiplied by 0.2 kg, multiplied by V one X squared plus one half, multiplied by 3.5 newtons per meter multiplied by 0 m squared is equal to one half multiplied by 0.2 kg, multiplied by 0 m per second squared plus one half multiplied by 3.5 newtons per meter multiplied by delta X two squared. And we're asked which problem statement is valid for this equation. We have four answer choices A through D we're gonna go through each of them. But the basic idea is that we have a ball of putty traveling and it's gonna interact with another ball of pike. When we have this interaction, we have some options and those balls could stick together, they could bounce off one of them. After that collision, they are going to compress a horizontal spring with a particular spring constant. And we want to find out how far the spring compresses. OK. So these are the basics of the, of the question. We need to figure out what, what kind of details are accurate kind of details. Give us this equation. So let's just work through and think through this problem, we're gonna come back to the answer choices at the end and kind of walk through them and figure out which one works and why the other ones don't work. So if we think about after the collision, if we wanna figure out how much the spring compresses, so let's think about after the collision, we're gonna have a conservation of mechanical energy. So the mechanical energy at time 0.1 emac one will be equal to the mechanical energy at time 0.2. And again, time 0.1 is going to be after the collision immediately after the collision. And time 0.2 is going to be when we have maximum compression. So I write that just so we remember now our mechanical energy, we're gonna have the kinetic energy and the potential energy. In this case, we have a horizontal spring. So we don't have to worry about any gravitational potential energy. We're only gonna have spring potential energy. And so we have K one plus us one. So kinetic energy one plus the spring potential energy one is equal to K two plus us two. Now recall that our kinetic energy is given by one half MV squared. Our spring potential energy is given by one half KX squared. So this equation is gonna become one half M V one squared plus one half kx one squared. That's gonna be equal to one half MV two squared plus one half kx two squared. All right, great. Now, this equation is starting to look like the same form as the equation we were given. So that's great news. Let's think about some of these terms. And let's write these compressions as delta X one and delta X two to kind of match with what we were given in these problems. OK? No, immediately after the collision, the spring has no compression, these balls have not struck that spring yet. So delta X one is zero. OK. So this term is zero. Delta X two is what we are looking for. OK. All right. So let's, let's start there. We have our one half term. OK. We have this spring constant if we have a spring constant of 3.5 newtons per meter. OK. So this is our K value and we know that delta X one is 0 m. OK? So that looks good that matches that equation. Now, if we compare again, the equation we were given, we can see that the mass used in the kinetic energy on both sides of our equation is 0.2 kg, we have 100 g balls. So in order for the mass to be 0.2 kg, they must have stuck together and be moving together with the same speed. OK. So because we have 0.2 kg, the balls must have stuck together right now. We don't have anything else in this equation. This is a conservation of mechanical energy. So we have no net external forces. That means that there we, there's no friction acting. OK. All right. So let's go through our answer choices now that we have kind of an idea of what's going on. So option A we have 100 g ball of putty traveling at 3 m per second. It strikes and sticks to another 100 g ball of putty at rest on a frictionless surface. The combined balls compress a horizontal spring with a spring constant of 3.5 newtons per meter. The other end of the spring is firmly anchored to a fixed post on the surface. How far will the spring compress? Well, that sounds like exactly what we have going on here. OK. We have the ball sticking together, we have a frictionless surface, we have a horizontal spring, we have the right value of the spring constant and we're looking for the compression. Ok? So option A looks good. What about option B? Well, option B has that the first ball of putty strikes and bounces off the other. Ok. So because it bounces off, we're not gonna get that total mass in the equation like we have. OK. So if they bounce off, then we don't have that total mass moving at the same speed. And so option B, we can eliminate option C and option C also has the ball sticking together. So that looks good, but it says that we have a rough surface with friction and we have the conservation of energy. That's how we got this equation. What that means is there are no net external forces so we can't have friction. So we can eliminate option C because of that option D we have a ball of putty at rest next to another ball of putty. They're pushed together by a horizontal spring with a particular spring constant. OK. This is not what we have going on. And we have the balls moving initially, they have an initial speed and no compression at the beginning and then afterwards we have some compression um and the balls come to rest, they, they come to rest when that compression is at its maximum momentarily as they change direction. And so option D is not correct either. And so as we thought, when we first read it, option A is going to be the correct answer. In this case, that is the correct problem statement to give us the equation. Thanks everyone for watching. I hope this video helped see you in the next one.